This course explains how to analyze circuits that have direct current (DC) current or voltage sources. A DC source is one that is constant. Circuits with resistors, capacitors, and inductors are covered, both analytically and experimentally. Some practical applications in sensors are demonstrated.

Преподаватели

Dr. Bonnie H. Ferri

Professor

Dr. Joyelle Harris

Director of the Engineering for Social Innovation Center

Dr Mary Ann Weitnauer

Текст видео

The topic of this problem is mesh and nodal analysis. And we're working with circuits with independent sources. The problem is to determine I sub zero in the circuit shown below using both mesh analysis and nodal analysis. When we're doing the nodal analysis, we'll use the concept of the super node. So here we go. We have the circuit below as three independent sources. Two of them are current sources. One of them is a voltage source. I sub zero is the current that flows through the two or the one-kilo ohm resistor at the center bottom of the circuit. So we're looking for I sub zero. So if we're going to use nodal analysis, we'll start with nodal analysis and solve it first using nodal analysis. And the first thing we do when we're doing nodal analysis is we assign the nodes in the circuit. So our first node would be here. A second node in the center of the circuit and a third node, my right-hand side of the circuit. The second thing we do is we sum the currents either into or out of those nodes. It's our choice as to which we choose. In this case, let's sum the currents out of the nodes. Starting with node one, and we know we use Kirchhoff's current law which states that the sum of the currents into any node at any given instant in time is equal to zero. So if we look at node one, we know that we have two milliamps flowing out to the bottom branch. We have four milliamps flowing out to the top branch, and we have a certain current which is flowing from the node one to node two, through the two-kilo ohm resistor. And we know that current is going to be 2K for the resistance and V1 minus V2, for the voltage. So, the current is V sub 1 minus V sub 2, divided by 2K, and that's equal to zero. So that's the sum of the currents into or out of node one. Now, if we look at node two and node three, we see that we have this voltage source between node two and node three. And we really don't have a way of determining what the current is through the voltage source immediately. Unless we want to assign another variable, maybe I12 volts with a current that flows through it. But that would then give us a third or a fourth unknown in our equations. Right now, we have unknown V sub one and V sub two. As we continue to write equations, we're going to have an unknown V sub three. So we're going to need three equations for those three unknowns. And if we add another one for the current through the 12 volt source, and that would be our fourth equation. We wouldn't be able to come up with a fourth independent equation to solve the problem. So the next thing we notice is that, since we have this voltage source, which is floating between node two and node three. Then we have to use the concept of a super node to solve the problem. And the way we do that is we kind of block off this voltage source like this. And we'll sum all the currents out of what we call a super node, which is the node which encompasses what we assigned as node two and node three, and the 12-volt voltage source. So if we sum the currents out of the super node, it's going to be the current right to left, through the two-kilo ohm resistor. Top to bottom through the one -kilo ohm resistor, top to bottom through the two kilo-ohm resistor. And then we also have the four milliamps which is flowing in. So we have a minus four milliamp contribution flowing out from this path. So it's going to have four different components to it. So this is the second equation. It's our super node equation where we have, first of all, the current flowing right to left through the two kilohm resistor. Which we know is V2 minus V1, over 2K, plus the current flowing top to bottom through the one kilo ohm resistor. It's equal to I sub zero, by the way. And that is going to be V2 minus 0, divided by 1K, because we know that the bottom of our circuit is a ground node. And we have the current down through the two kilo ohm resistor on the right-hand of our circuit, V sub 3 minus 0, over 2K. And the current which is flowing up which is minus four milliamps, so it's minus four milliamps. And that's all the currents flowing out of the super node. And so again we have an equation that has three unknowns V1, V2, and V3. We have two independent equations at this point. So we need one more equation relates V1, V2, and V3. And our third equations comes from the constraining equation of our super node, that is V sub 3 minus V sub 2, is equal to 12 volts. Okay, so now we have the three equations and three unknowns. We know that we get from nodal analysis, we get nodal voltages. So the result of our analysis is going to give us V1, V2 and V3. If we want to find something like I sub zero, we an find it easily once we know what V sub two is. If we know the voltage at this node, I sub zero simply that voltage divided by one kilo ohm to get that current. So, first thing we notice is that if we solve these equations, we get a V sub 2 equal to minus 16 over 3 volts. And we know that I sub zero is going to be equal to minus 5.33 milliamps. So that's our nodal analysis for this problem. What we're going to do is we're going to also solve this using mesh analysis. If we use mesh analysis, we'll see that we get the same answer for I sub zero is going to be equal to minus 5.33 milliamps. So let's do that, the first thing we do when we're solving a problem using mesh analysis is we assign meshes and mesh currents. So let's start with doing that. We'll call this mesh one, with mesh current I sub one. We'll call this mesh two, with mesh current I sub two. And we'll call this mesh three, with mesh current I sub three. So as we go around and look at these different meshes. We know that when we're doing mesh analysis, that we're looking for mesh currents. It's a complement to what we did in nodal analysis, where we looked for nodal voltages. So if we look at mesh one, we see that we can determine sub one directly from our mesh. That is, I sub 1 is equal to minus 2 milliamps. Our second equation for our second loop gives us a similar way to find I sub two. I sub two in this case is the only current that's flowing through this four-milliamp source at the top of our circuit. So we know I sub two is equal to four milliamps. So that's two out of our three mesh currents. The third one that we need and in fact, we need it to find I sub zero, is we're going to need to find I sub three. So we can sum up the voltages around this third loop to find I sub three. So if we start the lower left hand corner of this loop and go up clockwise, we first encounter the one-kilo ohm resistor. The voltage drop across the one-kilo ohm resistor is 1K times I sub 3, which is flowing in the same direction that we're summing our voltages minus I1, which is flowing the opposite direction. We continue up around our mesh or loop, we encounter the 12-volt source. In fact, we encounter the negative polarity of that 12-volt source. So it's minus 12. We continue and we encountered the two-kilo ohm resistor on the right-hand side of our circuit as we complete our mesh. And the voltage drop for that element is going to be 2K times I sub 3, and that's equal to zero. So we have a third equation which is independent of our first two and it tells us that, if we know I1 and I2, we can find I sub three. And I sub three in this case, comes out to be 3.33 milliamps for I sub three. Now, if we want to find I sub zero, I sub zero is going to be equal to I1 minus I sub 3. This is I sub one. And so that's equal to I sub one, which was minus two milliamps, minus I sub three, which was 3.33 milliamps. So we get the same answer for I sub zero. We have an I sub zero equal to minus 5.33 milliamps, just as we did for nodal analysis.